I'm trying to solve the following problem from Linear Algebra Done Right, by Sheldon Axler.
Prove or give a counterexample: if $v_1, \dots , v_m \in V$, then $$\sum_{j = 1}^{m}\sum_{k = 1}^{m}\langle v_j, v_k \rangle \geq 0.$$
I don't know if the statement is true, so all I've done so far is try to manipulate the expression: $$\sum_{j = 1}^{m}\sum_{k = 1}^{m}\langle v_j, v_k \rangle = \sum_{j = k}^{}\sum_{k = 1}^{m} \langle v_j, v_k \rangle + \sum_{j \neq k}^{}\sum_{k = 1}^{m}\langle v_j, v_k \rangle = \sum_{j = 1}^{m} \langle v_j, v_j \rangle + 2 \sum_{k > j} \Re(\langle v_j, v_k \rangle),$$ whose first term is nonnegative, by defintion of inner product. However, I can't say anything about the second term. Any help would be appreciated.
The trick here comes down to the linearity of the dot-product: $$ \sum_{j = 1}^{m}\sum_{k = 1}^{m}\left\langle v_j, v_k \right\rangle = \sum_{j = 1}^{m}\left\langle v_j, \sum_{k = 1}^{m}v_k \right\rangle = \left\langle\sum_{j = 1}^{m} v_j, \sum_{k = 1}^{m}v_k \right\rangle $$ That is, the expression can be written as $\langle w,w\rangle$ where $w = \sum_{j = 1}^{m} v_j$.